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155
7Phase-Shifting and Zig-Zag Transformers
7.1 Introduction
Phase-shifting transformers are used in power systems to help
control power flow and line losses. They shift the input voltages
and currents by an angle that can be adjusted using a tap changer.
They operate by adding a 90 volt-age to the input voltage, that is,
in quadrature. For three-phase transformers, the quadrature voltage
to be added to a given phase voltage can be derived by
interconnecting the other phases, which can be done in many ways;
this gives rise to a large number of configurations for these
transformers. We will deal with only a few common types of
configurations in this chapter. The phase-shifting capability can
be combined with voltage magnitude control in the same transformer.
This results in a more complex unit involving two sets of tap
changers; however, we will not discuss this in this chapter.
The use of phase-shift transformers can be illustrated by
considering the feeding of a common load from two voltage sources
that are out of phase with each other, as shown in Figure 7.1.
Solving for the currents, we get
I V V I V V1 12
2 2 21
1=( )
=+( )
Z Z
KZK
Z Z
KZK
+ L L L L, (7.1)
where K = Z1Z2 + Z1ZL + Z2ZL. The current into the load, IL,
is
I I I V VL = + = +1 2 1 2 2 1ZK
ZK
(7.2)
and the voltage across the load, VL, is
V V I V VLL= = +( )1 1 1 1 2 2 1Z ZK Z Z (7.3)
Thus, the complex power delivered to the load is
V I V VL LL* = +Z
KZ Z2 1 2 2 1
2 (7.4)
where * denotes complex conjugation.
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156 Transformer Design Principles
Consider a case in which Z1 = Z2 = Z, V1 = V, V2 = Vej. Then,
Equation 7.4 becomes
V IL LL* ( cos )= +2 1
2 2
2
Z Z VK
(7.5)
Thus, the maximum power is transferred when = 0. In this case, a
phase-shifting transformer can be used to adjust the phase of V2 so
that it equals that of V1.
In modern interconnected power systems the need for these
devices has increased. Other methods for introducing a quadrature
voltage are being developed, for instance, by means of power
electronic circuits in conjunc-tion with ACDC converters, which can
act much faster than on-load tap changers. However they are
presently more costly than phase-shifting transformers and are used
primarily when response time is important. (Electronic on-load tap
changers are also being developed for fast response-time
applications.)
In this chapter, we will develop a circuit model description for
three com-mon types of phase-shifting transformers. This is useful
to understand the regulation behavior of such devices, that is, how
much the output voltage magnitude and phase change are when the
unit is loaded as compared with that of the no-load voltage output.
In the process, we will also find that the phase angle depends on
the tap position, a relationship that can be nonlinear. In addition
to the positive-sequence circuit model, which describes normal
operations, we will also determine the negative- and zero-sequence
circuit models for use in short-circuit fault current analysis,
which is carried out for the standard types of faults for two
phase-shifting transformers.
Very little has been published on phase-shifting transformers in
the open literature beyond general interconnection diagrams and how
they are used in specific power grids. References that emphasize
the basic principles include [Hob39, Cle39, Kra98, Wes64].
V1 V2
Z2 Z1
ZL
I2 I1
Figure 7.1Two possibly out-of-phase sources feeding a common
load.
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Phase-Shifting and Zig-Zag Transformers 157
7.2 Basic Principles
The basic principles have been discussed in Chapters 3 through
6, but we will repeat some of them here for easier reference. We
neglect the exciting current and model the individual phases in
terms of their leakage impedances. For a two-winding phase, we use
the circuit model shown in Figure 7.2. Z12 is the two-winding
leakage impedance, referred to as side 1. Most of the develop-ment
described here is carried out in terms of impedances in Ohms.
Because of the differences in per-unit bases for the input and
winding quantities, per-unit quantities are not as convenient in
this analysis. We indicate the appro-priate places where the
per-unit quantities might prove useful. We assume that positive
currents flow into their respective windings. With N1 equal to the
number of turns on side 1 and N2, the number of turns on side 2,
the ideal transformer voltages satisfy
E12
1
2E= N
N (7.6)
and the currents satisfy
II
1
2
2
1
= NN
(7.7)
For three windings per phase, we use the model shown in Figure
7.3. In this case, the ideal transformer voltages satisfy
E11
2
2
3
3N N N= =E E (7.8)
and the currents satisfy
N N N1 1 2 2 3 3 0I I+ + =I (7.9)
V1
Z12
E1 E2
I2I1 N1:N2
Figure 7.2Model of a two-winding transformer phase.
2010 by Taylor and Francis Group, LLC
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158 Transformer Design Principles
The single-winding impedances are given in terms of the
two-winding impedances by
Z Z ZNN
Z
ZNN
1 12 131
2
2
23
22
12
12
= +
=11
2
121
2
2
23 13
31
+
=
ZNN
Z Z
Z22
3
1
2
131
2
2
23 12NN
ZNN
Z Z
+
(7.10)
Here, the first subscript refers to the winding for which the
two-winding impedance is measured and the second subscript refers
to the winding that will be shorted when measuring the impedance.
To refer impedances to the opposite winding, we use
V1
Z1
E1
I1 N1
V2
Z2
E2
I2 N2
V3
Z1
E3
I1 N3
Figure 7.3Model of a three-winding transformer phase. The
vertical lines connecting the circuits repre-sent the common
core.
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Phase-Shifting and Zig-Zag Transformers 159
ZNN
Ziji
jji=
2
(7.11)
For a three-phase system, the positive-sequence quantities
correspond to the order of the unit phasors shown in Figure 7.4. We
let
= = +e j j120 12
32
(7.12)
Then, the ordering is 1, 2, and . This corre-sponds to the phase
ordering a, b, and c for the windings and 1, 2, and 3 for the
terminals. Negative-sequence ordering is 1, , and 2, which
corre-sponds to the phase ordering a, c, and b or 1, 3, and 2 for
the windings or terminals, respectively. This is obtained by
interchanging two phases. Note that
2 240 120 12
32
= = = = e ej j j* (7.13)
The zero-sequence quantities are all in phase with each other.We
will choose interconnections to produce a positive phase shift at a
posi-
tive tap setting. By interchanging two phases, we can produce a
negative phase shift at a positive tap setting. Interchanging two
phases is equivalent to imputing a negative-sequence set of
voltages. This implies that negative-sequence circuits have the
opposite phase shift in relation to positive-sequence circuits.
Zero-sequence circuits have zero phase shifts. By positive phase
shift, we mean that output voltages and currents lead the input
voltages and currents, that is, output quantities are rotated
counterclockwise on a phasor diagram relative to the input
quantities.
7.3 Squashed Delta Phase-Shifting Transformer
One of the simplest phase shifters to analyze is the squashed
delta configu-ration shown in Figure 7.5, where S denotes the
source or input quantities and L indicates the load or output
quantities. The input and output set of voltages and the
corresponding currents form a balanced positive-sequence set. In
this study, we take the S currents as positive into the terminals
and the L currents as positive out of the terminals. The input and
output voltage
b phase
c phase
1
2
j
a phase
Figure 7.4Positive-sequence unit phasors.
2010 by Taylor and Francis Group, LLC
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160 Transformer Design Principles
phasor diagram is shown in Figure 7.5b. These are the voltages
to ground; the currents form a similar set but are not shown.
Similarly, the internal volt-ages and corresponding currents form a
positive-sequence set, as shown in Figure 7.5c for the voltages.
Note that Figures 7.5b and 7.5c can be rotated relative to each
other if shown on a common phasor diagram. Information on their
relative orientation is not contained in the figure although the
phase order is consistent between the figures. This applies to the
corresponding current phasors as well. a, a, and so on refer to
windings on the same leg, with the prime labeling the tapped
winding. The two-winding impedances, that is, Zaa, and so on, will
be referred to the unprimed coil. Because all these impedances are
the same by symmetry, we need only one symbol. This transformer can
be designed with a single three-phase core and is generi-cally
referred to as a single-core design. It is also best adapted to
phase-angle shifts in one direction.
Using the two-winding phase model described in the Section 7.2,
adapting it to the present labeling scheme, and concentrating on
one inputoutput pair, we can write
VS L1 a S1 L c aa c
S1 c a L1,1 2 = = += +
V E V V I E
I I I I
, Z
== +I Ia b (7.14)
VS1VL1
VS3
VL3
VS2
VL2
Ea
Eb
Ec
Eb
Ec
Ea
IS1 IL1
IS3
IL3IS2
IL2
Ia
Ib
Ia
IbIc
Ic
(a)
VS1
VL1
VL2
VS2
VS3
VL3
(b)
Ea
Ea
Ec
Ec
Eb Eb
(c)
Figure 7.5Squashed delta configuration, with the phase
quantities labeled: (a) circuit, (b) input and output phasors, and
(c) internal phasors. The ideal transformer voltages, the Es,
increase in the opposite direction of the assumed current flow. is
a positive phase-shift angle.
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Phase-Shifting and Zig-Zag Transformers 161
Note that, in Equation 7.14, the two-winding leakage impedance
is assigned to the unprimed side of the transformer. We also have
the following trans-former relations:
E E E E
I I I
= = =
= =
aa
aa a c
aa
aa
eNN n nNN
n
j1 120
aa ce= n j120 I (7.15)
where we define n = Na/Na, the turns ratio. We also have the
following relationships:
V V V V
V V E
S S2 S1 S1
S2 L2 b
e
e
12 30
1
1 3 = ( ) =
= =
j
j 220120
=E Ea c
e j
n
(7.16)
A detailed solution of these equations is essential because we
will need some of the intermediate results later. We assume that
VS1 and IS1 are given. From Equations 7.14 and 7.15, we get
I I I Ic S1 a See
e=
=
11 1120
120
120nn
njj
j, 11 120=
nn je S1
I (7.17)
We also have
I I I I IL1 a b a ceee
= + = + =
j
j
j
nn
120120
120
=
=+
I IS1 S1e
tan
j
n
2 32 1
1
where
I I I I IL1 a b a ceee
= + = + =
j
j
j
nn
120120
120
=
=+
I IS1 S1e
tan
j
n
2 32 1
1 (7.18)
Because n is determined by the tap position, this shows that is
a nonlinear function of the tap position, assuming the taps are
evenly spaced.
Adding Equation 7.16 and using Equation 7.14, we get
V V V E I ES1 L2 S1 c c aa cee = + = +
3 30120
jj
nZ
Solving for Ec and using Equation 7.17, we get
E Vc S1aae
ee
e=
+
(
3 30120
120
120
nn
n Z
n
j
j
j
j ))2IS1 (7.19)
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162 Transformer Design Principles
Combining this equation with the first expressions of Equations
7.14 and 7.15, we get
V V IL1 S1aa
S1 e= + +
Zn n
j2 1
(7.20)
where is given by Equation 7.18. Although the current shifts by
in all cases, in general, the voltage shifts by only under no load
conditions.
The circuit model suggested by Equation 7.20 is shown in Figure
7.6. By symmetry, this applies to all three phases with appropriate
labeling. The equivalent impedance shown in Figure 7.6 is given
by
ZZ
n neqaa=
+ +
2 1 (7.21)
Figure 7.6 is a positive-sequence circuit model. The
negative-sequence model is obtained simply by changing to , with
Zeq unchanged. Note that Zeq depends on the tap setting.
Although the input power per phase is Pin = VS1IS1*, the
transformed or winding power per phase Pwdg is relatively less.
Ignoring impedance drops, from Equations 7.17 and 7.19, we have
P jn
n nPwdg c c in= = + +
E I*3
12 (7.22)
where the factor j shows that the transformed power is in
quadrature with the input power. As a numerical example, let = 30.
Then, from Equations 7.18 and 7.22, we have
tan wdg in2
32 1
2 732 0 4227 = +
= =n
n P j P. . (7.23)
VS1
Zeq
VL1
IL1IS1 : 1 ej
Figure 7.6Circuit model of one phase of a phase-shifting
transformer for a positive sequence. For a nega-tive sequence,
change to .
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Phase-Shifting and Zig-Zag Transformers 163
We can express Equation 7.20 in per-unit terms; however, because
of the difference between the input power and voltage base and that
of the wind-ings, we must be careful to specify the base used. As
usual, we take the power base to be the terminal power base. This
is the terminal power per phase. Thus, the rated input power base
is Pb,in; the rated input voltage base is Vb,in; and the rated
input current base that can be derived from the power and voltage
base is Ib,in = Pb,in/Vb,in. Similarly, we can derive the input
imped-ance base from the power and voltage base, Zb,in =
Vb,in/Ib,in = (Vb,in)2/Pb,in. Because the transformation ratio is
1:1 in terms of magnitude, the output base values are the same as
the input base values. Thus, Equation 7.20 can be written as
follows:
V V IL1b,in
S1
b,ineq
S1
b,in
b,in
b,inV VZ
IIV
=
=
e S1
b,in
eq
b,in
S1
b,in
j
V
Z
Z I V I
=
e
eL1 S1 eq S
j
jz
v v i 1
(7.24)
or
V V IL1b,in
S1
b,ineq
S1
b,in
b,in
b,inV VZ
IIV
=
=
e S1
b,in
eq
b,in
S1
b,in
j
V
Z
Z I V I
=
e
eL1 S1 eq S
j
jz
v v i 1
where lowercase letters are used to denote the per-unit
quantities.Assuming the rated input power per phase to be the
common power base,
the winding and input impedance bases are proportional to the
square of the respective voltage magnitudes. Using Equation 7.19
and ignoring impedance drops, we have
Z
Zn
n nb,wdg
b,in
c
S1
= =+ +
EV
22
2
31
(7.25)
Thus, using Equation 7.21, zeq can be expressed as
zZ
Z n n
Z
ZZ
Zeqeq
b,in
b,wdg
b,in
aa
b
= =+ +
1
12 ,,wdgaa
= + +( )
31
2
2 2
n
n nz (7.26)
where zeq is on an input base and zaa is on a winding base. Note
that the winding base depends on the turns ratio, as indicated in
Equation 7.25. Thus, this base is perhaps most useful for design
purposes when n refers to the maximum phase angle. At the other
extreme, with = 0, we have n = , so that Zeq = zeq = 0. In this
case, the input is directly connected to the output, bypassing the
coils.
7.3.1 Zero-Sequence Circuit Model
The zero-sequence circuit model may be derived with reference to
Figure 7.5, assuming all quantities are of zero sequence. Thus,
Figures 7.5b and 7.5c
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164 Transformer Design Principles
should be replaced by diagrams with all phasors in parallel.
Using a zero subscript for zero-sequence quantities, we can
write
V V E V V IS1,0 L1,0 a ,0 S L2,0 c,0 aa , c = = + , ,1 0 0Z E
,,0
S1,0 c,0 a ,0 L1,0 a ,0 b,0I I I I I I= + = + , (7.27)
and
E E E E
I I
= = =
=
aa
aa a c
aa
aa
, , , ,
,
0 0 0 0
0
1 1NN n nNN ,, , ,0 0 0
= = n nI Ia c (7.28)
Because Ib,0 = Ic,0, Equation 7.27 shows that
I IL1,0 S= 1 0, (7.29)
so the current is not phase-shifted. We also have
V V V V E E ES S S L b ,0 b c1 0 2 0 2 0 2 0 001 1
, , , , ,, = = = = n n ,,0 (7.30)
Solving the above zero-sequence equations, we get
I E I Ic,0S1,0
c,0 c,0 aa ,0 S= =
=
( )I
nn
nZ
nn1 1 1 2
, 11,0 aa ,Z 0 (7.31)
and
V V IL1,0 S1,0aa ,0
S1,0= ( )Z
n1 2 (7.32)
The circuit model for Equation 7.32 is shown in Figure 7.7,
where we define
ZZ
neq,0
aa ,0=( )
1 2 (7.33)
Zeq,0 has a different dependence on n compared to the
positive-sequence circuit and becomes infinite when n = 1. This is
reasonable because, from Equation 7.28, Ia,0 = Ic,0 when n = 1;
therefore, from Equation 7.27, IS1,0 = 0. Thus, no zero-sequence
current can flow into the squashed delta transformer when n = 1.
Internal current can, however, circulate around the delta. From
Equation 7.18, = 60 when n = 1.
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Phase-Shifting and Zig-Zag Transformers 165
Equation 7.32 can be rewritten in per-unit terms referring to
the same input base as used for the positive sequence:
v v iL1,0 S1,0 eq S1,0= z ,0 (7.34)
Using the positive-sequence winding base for the two-winding
zero-sequence impedance, Zaa,0, we can write
zn
n n nzeq,0 aa ,0= + +
31 1
2
2 2( )( ) (7.35)
This expression is primarily useful in designs in which the
winding base is used for the maximum phase-shift angle.
We will postpone both a discussion of the regulation effects and
calcula-tions of the short-circuit current until other types of
phase-shifting trans-formers are treated because the positive-,
negative-, and zero-sequence circuit diagrams for all the cases
treated will have the same appearance, although Zeq or zeq and
differ among the various types.
7.4 Standard Delta Phase-Shifting Transformer
As opposed to the squashed delta design, the standard delta
design uses an unsquashed delta winding but is still a single-core
design. The connec-tion diagram is given in Figure 7.8. The tapped
windings are on the same core as the corresponding parallel
windings on the delta. The taps are sym-metrically placed with
respect to the point of contact at the delta vertex. This ensures
that there is no change in the current or no-load voltage magnitude
from the input to the output. It also means that Ea = Ea, and so
on, for the other phases. Each phase really consists of three
windings: the two tap windings and the winding opposite and
parallel in Figure 7.8, where primes
VS1,0
Zeq,0
VL1,0
IS1,0 IL1,0
Figure 7.7Zero-sequence circuit model of one phase of a squashed
delta phase-shifting transformer.
2010 by Taylor and Francis Group, LLC
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166 Transformer Design Principles
and double primes are used to distinguish them. Thus, a
three-winding per-phase model is needed. Za, Za, and Za will be
used to label the single-winding impedances for phase a. Because
all the phases have equivalent impedances by symmetry, the same
designations will be used for the other phases as well. The same
remarks apply to the phasor diagrams as in the squashed delta case.
Although not shown, the current phasor diagrams have the same
sequence order as their corresponding voltage-phasor diagrams;
however, the two diagrams can be rotated relative to each
other.
(b)
(a)
(c)
VS1 VL1
VS3
V2
VL2
VL3
VL1VL3
VL2
EaEa
EcEb
Ec
IS1
VS2
VS3
V3
VS1
IS2
VS2
IL1V1
IS3IL2
IL3
Ia Ia
Ib
IaEa
IcIb
Ib
Eb
Ec
Eb
Ea
Ea
Eb
Ec
Eb
EaEc
Eb
EcIc
Ic
Figure 7.8Standard delta configuration, with the phase
quantities labeled: (a) circuit, (b) input and output phasors, and
(c) internal phasors. The ideal transformer voltages, the Es,
increase in the oppo-site direction of the assumed current flow.
The taps are symmetrically positioned.
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Phase-Shifting and Zig-Zag Transformers 167
In Figure 7.8a, V1, V2, and V3 designate the phasor voltages to
ground at the delta vertices. Thus, using Ea = Ea and so on, and Ia
= IS1; Ia = IL1, and so on, we have
V V I E V V I E
V V
S1 S a a L L a a = + = +
= 1 1 1 1 1
2 3
Z Z,
(
21 1
12
3 = = +
= =
)
( )
V V I E
I I I I I
j Za a a
S1 L c b aa a3= j I
(7.36)
In addition, we have the following transformer relations:
EE
I I I I I
a
a
a
a
a a a S a L a Sor
= =
+ + = + +
NN
n
N N N n1 1 10 IIL1 0=
(7.37)
where we define the turns ratio n as the ratio between the turns
in one of the delta windings to the turns in one of the tap
windings. Both tap wind-ings have the same number of turns. Solving
for Ia in the last equation in Equation 7.36 and inserting it into
the last equation in Equation 7.37, we get
I IL1 S1e with tan= =
j
n 2 31 (7.38)
We also have
I Ia S1=
2
13
n jn
(7.39)
From Equations 7.36 through 7.39, we get
E Va S a
a=
+
j
jn
j
jn
ZjZ3
13
3
13
21
33 13
1
n jn
IS (7.40)
and
V V IL1 S1 a aa
S1 e= + + +
Z ZZ
nj4
32 (7.41)
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168 Transformer Design Principles
with as given in Equation 7.38. This is the no-load phase-angle
shift. This conforms to the circuit model shown in Figure 7.6
with
Z Z ZZ
neq a aa= + ++
432
(7.42)
or in terms of two-winding impedances, using Equation 7.10:
Z Zn
Z Z n Zeq a a aa aa a a= + +
+ ( )
232
2 (7.43)
This is the positive-sequence circuit. Again, the
negative-sequence circuit is found by changing to without any
change in Zeq.
We again determine the winding power per phase Pwdg in terms of
the input power per phase Pin = VS1IS1*. Similarly, ignoring the
impedance drops, we get
P jn
nPwdg a a* in= = +
E I2 3
32 (7.44)
where j indicates that the transformed power is in quadrature
with the input power. Using trigonometric identities, we can show
that
P j Pwdg insin= (7.45)
with as given in Equation 7.38. (This relation does not hold for
the squashed delta design.) Thus, for = 30, |Pwdg| = 0.5 |Pin|.
In per-unit terms, based on input quantities, Equation 7.41 can
be cast in the form of Equation 7.24. Again, we need to determine
the relationship between the winding bases and the input base for
impedances. The two-winding impedances given above are referred to
by either the a or a winding. Their bases are related to the input
base by maintaining the power base the same as the input power per
phase:
ZEV
nn
Z Zb,a wdga
S1b,in b,a wd= = +
22
2
33
, gga
S1b,in= = +
EV n
Z2
2
33
(7.46)
Thus, in per-unit terms, Zeq in Equation 7.43 becomes
zn
nzn
z z zeq
a a aa aa a=+
+
+ 33
222 2
aa( )+( )
n2 3
(7.47)
Again, this is primarily useful for design purposes when n
refers to the maximum phase angle. At zero phase shift, n = ,
Equation 7.47 indicates that zeq = 0. At this tap position, the tap
windings are effectively out of the circuit, and the input is
directly connected to the output.
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 169
7.4.1 Zero-Sequence Circuit Model
The zero-sequence circuit model can be derived with reference to
Figure 7.8 but with all phasors taken to be zero sequence.
Rewriting Equations 7.36 and 7.37 accordingly and appending a zero
subscript, we get
V V I E V V IS1,0 S1,0 a ,0 a ,0 L L1, = + = 1 0 1 0 1 0, , ,,Z
00 a ,0 a ,0
a, a,0 a, S1
Z
Z +
= = +E
V V I E I2 0 3 0 0 00, , , ,,0 L1,0 c,0 b,0
a
a ,0
a
aa,
= =
= =
I I I
EE
I
0
0, ,NN
n n 00 0+ + =I IS1,0 L1,0
(7.48)
Solving, we find that
I IL1,0 S1,0= (7.49)
thus, the zero-sequence current undergoes no phase shift. We
also have
E I I Ia,0 a,0 a,0 a, S1,0,= = Z n02 (7.50)
and
V V IL1,0 S1,0 a ,0 a ,0 a,0 S1,0= + +
Z Z n
Z4
2 (7.51)
Thus, the voltage undergoes no phase shift at no-load. The
circuit model of Figure 7.7 applies here, with
Z Z Zn
Zeq a ,0 a ,0 a,0= + + 4
2 (7.52)
or, in terms of two winding impedances
Z Zn
Z Z n Zeq,0 a a ,0 aa ,0 aa ,0 a a ,0= + + 2
22(( ) (7.53)
On a per-unit basis, using rated input quantities, Equation 7.51
has the same appearance as Equation 7.34. The equivalent per-unit
impedance is given by
zn
z z zeq,0 aa ,0 aa ,0 a a ,0= +
+( )
33
22
(7.54)
Again, this is primarily useful when n refers to the maximum
phase angle shift. At zero phase shift, the output is directly
connected to the input as was the case for the positive
sequence.
2010 by Taylor and Francis Group, LLC
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170 Transformer Design Principles
7.5 Two-Core Phase-Shifting Transformer
Phase shifters for large power applications are often designed
as two unitsthe series unit and the excitor unit, each with its own
core and associated coils. Depending on their size, the two units
can be housed inside the same tank or in separate tanks. This
construction is largely dictated by tap-changer limitations. A
commonly used circuit diagram is shown in Figure 7.9. The phasor
diagrams refer to the positive-sequence quantities and, although
the phase ordering is consistent among the dia-grams, their
relative orientation is not specified. Currents have the same phase
ordering as their associated voltages. The series unit uses a
three-winding model, whereas the excitor unit uses a two-winding
model. We have used two different labeling schemes for the series
and excitor units. Subscript letters are used for the series
quantities, and subscript numbers are used for the excitor
quantities. Primes and double primes are used to distinguish
different windings associated with the same phase. Note that the
inputoutput coils in Figure 7.9 are a part of the series unit but
are attached to the excitor unit at their midpoints. The input and
output voltages are voltages to ground. We assume the input voltage
and current phasors are given.
Following an analysis similar to that of the Section 7.4 for the
series unit, and using Ea = Ea, and so on, we can write
V V I E V V I E
E E
S S Z Z1 1 1 1 1 1
2 3
= + = +
a a L L a a ,
== ( ) = = += + =
2 1 11 1 1 1 1
3E E I EV I E I
j Z
Z
a a a1 , II I I I IS1 1 2 = L c a,
(7.55)
We also have the following transformer relations:
EE
EE
II
a
a
a
as
1e, ,
= = = = = =N
Nn
NN
nNN
1
1 1
1
1
1
1
+ + = + + =
n
N N N n
e
a a a S1 a L1 s a S1 L1orI I I I I I0 0
(7.56)
where we define the turns ratio of the series unit, ns, as the
ratio of the turns in a coil of the delta to the turns in the first
or the second half of the inputoutput winding, that is, from the
input to the midpoint or from the output to the midpoint. We have
also defined the excitor-winding ratio, ne, as the ratio of the
turns in the winding connected to the midpoint of the inputoutput
winding to the turns in the tapped winding. The latter ratio will
depend on the tap position. From Equations 7.55 and 7.56 and the
phasor diagrams, we get
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 171
I I I I I I = = = = = 2 120 1 120 1 1e e c a aj e jn ( ) 33
3
30
1
e a
ea
=
j
jn
I
I I
(7.57)
which implies
I I I I I I = = = = = 2 120 1 120 1 1e e c a aj e jn ( ) 33
3
30
1
e a
ea
=
j
jn
I
I I
(a)
(b)
(c)
(d)
Series unit
Excitor unit
Ec
Ea
Ea EaEc
Ec
Eb
E3
E2
E1
E1
E2
E3
Eb Eb
Ec Ib
Ic
Ia
V1
E1
E3
E3
E1
E2VL2
IL2Eb
V2IS2
EbVS2 VL3
IL3
V3
IS3Ec
VS3
Ec
E2
I1
I3
I3
I1
I2
I2
IL1VL1VS1
ISI
Eb
Ea
EaEa
VS3VL1
VS1
VL2
VS2
VL3
Figure 7.9Circuit diagram of a two-core phase-shifting
transformer: (a) circuit, (b) input and output pha-sors, (c)
internal phasors for series, and (d) internal phasors for excitor.
The ideal transformer voltages, the Es, increase in the opposite
direction of the assumed current flow. The inputoutput coil is
equally divided.
2010 by Taylor and Francis Group, LLC
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172 Transformer Design Principles
Substituting I1 from Equation 7.55 and Ia from Equation 7.56
into Equation 7.57, we get
ILe s
e s
1
13
13
=+
jn n
jn n
= =
I IS1 S1e s
e where tanjn n
2 31 (7.58)
Thus, in terms of the known input current, using Equations 7.55
and 7.56, we can write
I I1 1
23
13
=
jn n
jn n
e s
e s
S , II Ia
se s
S1=
2
13
n jn n
(7.59)
From the initial set of equations, we also get
E I E1 3= +( )j n Ze a a a (7.60)
Substituting Equations 7.57 and 7.60 into the expression for V1
in Equation 7.55, we find that
V I E1 112
13 3= +
+Z
nZ j
nea
ea (7.61)
Substituting this into the first relationship of Equation 7.55,
and using Equation 7.56 and the first equation of Equation 7.59, we
get
E Vae
e s
S1
s
=
+j
n
jn n
n3
13
6nn n
jn n
Zn
Ze s
e s
ea
( )
+ 2
2 11
2
13 3
+
jn
jn n
Z
3
13
e
e s
a
IS1
(7.62)
Adding the first two expressions in Equation 7.55 and using
Equation 7.62, we get
V VL1 S1 a ae s
ea= + + ( ) +
+
Z Zn n
Zn
Z12
3 3211
2
IS1 e j (7.63)
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Phase-Shifting and Zig-Zag Transformers 173
where is given in Equation 7.58. Equation 7.63 can be
represented with the same circuit model as Figure 7.6 by applying
the following relation:
Z Z Zn n
Zn
Zeq a ae s
e
3= + +
( ) +
+ 12
3211
2
aa
(7.64)
or, in terms of the two-winding impedances, we can express it
as
Z Zn n
Zn
Zeq a ae s
e2
aa6= +
( ) +
+ +
1232
11 ZZ n Zaa s a a ( )
2 (7.65)
Using Equations 7.59 and 7.60 and ignoring impedance drops, we
can show that the winding power per phase is the same for the
series and excitor units, that is,
P jn n
n nPwdg a a*
e s
e si= = = ( ) +
E I E I1 1 22 3
3*
nn (7.66)
where Pin = VS1IS1*. In terms of the phase shift given in
Equation 7.58, Equation 7.66 can be written as
P j Pwdg insin= (7.67)
similar to the case for the standard delta phase shifter.In
per-unit terms, keeping the power base constant at the rated
input
power, Equation 7.63 can be cast in the form of Equation 7.24.
To do this, we need to find the ratio of the impedance bases for
both the series and excitor windings to the input or terminal
impedance base. Using the above formu-las, we find that
Zn
n nZ
Z
b,awdga
S1
s2
e sb,in
b
= =( ) +
EV
2
2
3
3
,,a wdga
S1 e sb,in
b,1
= =
( ) +
EV
2
2
3
3n nZ
Z wwdg1
S1
e s
e sb,in= =
( )( ) +
EV
2 2
23
n n
n nZ
(7.68)
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174 Transformer Design Principles
Applying these relations, the per-unit equivalent impedance can
be written as
zn n
zn n
n neq
e s
a ae s
e s
=( ) +
+ ( )( ) +
3
3
4
32
2
2
+ + ( )
z z z z11
12 aa aa a a
(7.69)
The negative-sequence circuit has the same equivalent impedance,
but it has a phase-angle shift in the opposite direction to the
positive-sequence circuit.
7.5.1 Zero-Sequence Circuit Model
The zero-sequence circuit model is derived with reference to
Figure 7.9 by assuming that all quantities are zero sequence. Thus,
we simply rewrite the basic equations, appending a zero
subscript:
V V I E V V IS1,0 S1,0 a ,0 a ,0 L L1,, = + = 1 0 1 0 1 0, , ,Z
00 a ,0 a ,0
a,0 a,0 a,0
Z
Z
+ = = +
E
E E I E
V2 0 3 0
1
0, ,,, , , , , ,0 1 0 11 0 1 0 1 0 2 0= + = = I E I I I IZ ,
,S1,0 L1,0 II Ic a,0,0 0 =
(7.70)
and
EE
EE
Iaa
a
as e
,
,
,
,
,, ,00
1 0
1 0
1
1
1 0
= = = =NN
nNN
nII
I I I1 0
1
1
0 1 0 1 0 0,
, , ,
= =
+ + =
NN
n
N N N
e
a a a S a L orr ns a S LI I I, , ,0 1 0 1 0 0+ + =
(7.71)
Solving, we find that
I I I I I
E I
1 0 1 0 1 002
2
, , ,= = =
=
, ,S L a,0s
S1
a,0s
S1,
n
n 00 a,0,Z V1 0 1 0, ,= E
(7.72)
Notice that, even if both the Y windings of the excitor are
grounded, no zero-sequence current flows into the excitor because
the secondary current from the tap winding would have to flow into
the closed delta of the series unit, and this is not possible for
zero-sequence currents.
From the above formulas, we get
V V IL1,0 S1,0 a ,0 a ,0s
a,0 S1,= + +
Z Zn
Z4
2 00 (7.73)
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Phase-Shifting and Zig-Zag Transformers 175
Thus, we see that the zero-sequence current and no-load voltage
have no phase angle shift and the circuit of Figure 7.7 applies
with
Z Z Zn
Zeq,0 a ,0 a ,0s
a,0= + + 4
2 (7.74)
or, in terms of two-winding impedances
Z Zn
Z Z n Zeq,0 a a ,0s
aa aa s a a= + + ( ) 22 2 (7.75)
Using the same basis as for the positive sequence, the per-unit
version of this equation is
zn n
z z zeq,0e s
aa aa a a= ( ) + +( )
3
322 (7.76)
7.6 Regulation Effects
Because all the phase-shifting transformers examined here have
the same basic positive-sequence (as well as negative- and
zero-sequence) circuits (Figure 7.6), with different expressions
for Zeq or zeq and , the effect of a load on the output can be
studied in common. The relevant circuit model is shown in Figure
7.10.
We assume a balanced positive-sequence system. Since all phases
are identical, we exclude the phase subscript. We use per-unit
quantities for this development, which can be accomplished by
simply changing from upper- to lowercase letters and using a unit
turns ratio. The ideal transformer is included in the figure to
account for the phase shift. From the figure, we see that
v v i i iL S eq S L L S Le e= ( ) = = z z zj j (7.77)
iLiS 1:e j
vS
zeq
zLvL
Figure 7.10One phase of a phase-shifting transformer under load,
represented using per-unit quantities.
2010 by Taylor and Francis Group, LLC
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176 Transformer Design Principles
where zL is the per-unit load impedance. Solving for iS, we find
that
iv
SS
eq L
=+( )z z (7.78)
so that
v v vL Seq
eq L eq
L
e= +( )
=+
1
1
1
z
z z z
z
jS
e j (7.79)
Thus, any shift in the phase angle or magnitude from the no-load
conditions is due to a nonzero zeq/zL. Because zeq is almost
entirely inductive, a purely inductive or capacitive load will not
affect but will result only in a magni-tude change in the voltage.
On the contrary, a resistive or complex load will lead to shifts in
both magnitude and phase angle.
Under no-load (NL) conditions (zL = ), Equation 7.79 shows
that
v vL,NL Se= j (7.80)
Thus, taking ratios, we see that
v
vL
L,NL eq
L
=+
1
1z
z
(7.81)
Therefore, the presence of a load will lower the magnitude and
shift the phase, depending on the phase of the above ratio of
impedances. This analy-sis is similar to that given in Chapter 3,
except there we ignored the shift in phase. Because these
transformers are designed to shift the phase by a given amount
under no-load conditions, finding the additional shift in this
phase caused by transformer loading is important. Let zeq/zL = pej.
Then, Equation 7.81 becomes
v
v p p
jppL
L,NL
tansincose=
+ +
+
11 2 2
11
cos
(7.82)
As an example, let zeq be a 10% inductive leakage impedance and
zL be a 100% resistive load. Then, zeq/zL = 0.1ej90 and vL/vL,NL =
0.995 5.71, where indi-cates the angular dependence of the complex
number. This angle of nearly 6 will be subtracted from the phase
angle of the transformer output, so the behavior of the phase under
load is quite important. In the above example,
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 177
the magnitude will only be lowered by 0.5%. Because the
transformer imped-ance varies with the tap setting (or phase
shift), the angle shift due to loading also varies with the tap
setting.
7.7 Fault Current Analysis
The phase-shifting transformers discussed in this section are
all two- terminal transformers that have a single reactance, Zeq or
Zeq,0, for positive- or negative- and zero-sequence circuits,
respectively. We assume that the positive- and negative-sequence
reactances are the same and do not distin-guish them with
subscripts. Thus, the theory developed in Chapter 6 can be applied
to these phase-shifting transformers with the recognition that
neg-ative-sequence currents have a phase shift opposite of the
positive-sequence currents. We retain the notations source, S, and
load, L, used above instead of H and X. We assume the fault is on
the L or output terminal. The fault cur-rents at the fault position
are the same for the various fault types, as shown in Chapter
6.
The faults of interest are as listed below:
1. Three-phase line-to-ground fault 2. Single-phase
line-to-ground fault 3. Line-to-line fault 4. Double line-to-ground
fault
We will summarize the fault currents for the above types here.
We will use per-unit notation.
For fault type 1, we have
iv
zi ia
pfa2 a0,1
1
0= = = (7.83)
For fault type 2, we assume that the a-phase line is shorted, so
that
iv
z z zi ia1
pfa2 a0= + +( ) = =0 1 2 (7.84)
For fault type 3, assuming that the b and c lines are shorted
together, we have
iv
z zi ia1
pfa2 a,= +( ) = =1 2 0
0 (7.85)
2010 by Taylor and Francis Group, LLC
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178 Transformer Design Principles
For fault type 4, assuming that the b and c lines are shorted to
ground, we get
i vz z
z z z z z z
i vz
z
a1 pf
a2 pf
= ++ +
=
0 2
0 1 0 2 1 2
0
00 1 0 2 1 2
2
0 1 0 2 1
z z z z z
i vz
z z z z z z
+ +
= + +a0 pf 22
(7.86)
Assuming equal positive and negative reactances for the
transformer and the systems, we have the following expressions for
the Thevenin impedances:
z zz z z
z z zz
z z1 2 0
0 0= =+( )
+ +=SL eq SS
eq SS SL
SL eq,
, , ++( )+ +
z
z z zSS
eq SS SL
,
, , ,
0
0 0 0
(7.87)
The per-unit currents feeding the fault from the transformer
side are given by
i iz
z zi i
zz zeq a eq SS
eq2 a2eq SS
,1 11 1=+
= +
=+
i i
zz zeq0 a0 eq,0 SS,0
0 (7.88)
Figure 7.11 shows the terminal circuit diagrams for the three
sequences, cov-ering all the fault types. For simplicity, we assume
that no current flows in the prefault condition. These can always
be added later.
As shown in Figure 7.11, the per-unit currents flowing out of
the trans-former into the fault are on the load side of the
transformer and are given by Equation 7.88 in terms of the fault
sequence currents. Thus, in terms of finding the fault currents in
the individual windings, we need to find their expres-sions in
terms of the load currents. However, in a phase-shifting
transformer, positive- and negative-sequence currents experience
opposite phase shifts. This makes the analysis of fault currents
within these transformers quite dif-ferent from the fault currents
in a standard two-winding transformer.
We have already considered the currents in the windings of a
phase-shifting transformer when we derived their equivalent
impedance for positive-, negative-, and zero-sequence circuits.
Now, we use those results and the fact that negative-sequence
currents within a phase-shifting trans-former can be obtained from
the positive-sequence currents by changing their phase shift to its
negative. Note that ieq in Equation 7.88 equals iS1 in per-unit
terms. We now apply the fault formulas from Equation 7.88 to the
phase-shifting transformers described in Sections 7.3 through 7.5.
Because the per-unit notation is used, we need to express the
formulas for the currents
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 179
in per-unit terms. Considering the lengths of the formulas, we
only present the results for the winding-sequence currents.
7.7.1 Squashed Delta Fault Currents
From Equation 7.17, we see that
I I I =
=
a S1 L1e ee
nn
nnj j
j120 120
(7.89)
where
= +
23
2 11tan
n (7.90)
Using Equation 7.90, Equation 7.89 can be manipulated to the
following form:
I I=
+ +a L1e
nn n
j
22
1
(7.91)
(a)
(b)
(c)
ZSS
ZSS
ZSS,0
Zeq
Zeq
Zeq,0
ZSL
ZSL,0
ZSL
eSLia1
ia2
ia0
iL1
iSL2
iSL0
ieq1
ieq2
ieq0
Va1
Va2
Va0
eSS
Figure 7.11Sequence circuits showing system impedances and
transformer equivalent impedances. The fault is on the load-side
terminal: (a) positive sequence, (b) negative sequence, and (c)
zero sequence.
2010 by Taylor and Francis Group, LLC
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180 Transformer Design Principles
To convert this expression to the per-unit form, we need the
ratios of the cur-rent bases. These are the reciprocals of the
ratios of the voltage bases, assum-ing the same power level for
both bases. This is given by
II
VV n n
b,L1
b,a
b,a
b,L1
= =+ +3
12 (7.92)
Using this relation, we can convert Equation 7.91 to the
per-unit form:
i i=
+ +a L1e
nn n
j312
2
(7.93)
Because this is a two-winding transformer, ia = ia.For the
zero-sequence currents, from Equations 7.28 and 7.31, we have
I I I = = a ,0 c,0 L1n
nn 1
(7.94)
Note that the zero-sequence current is the same in all the
phases. Using the base transformation given in Equation 7.92, this
can be converted to per-unit form:
i i = + +a ,0 L1n
n n n3
1 12( ) (7.95)
We also have ia0 = ia0.Using Equation 7.88 and the fact that
negative-sequence currents have
an opposite phase shift from positive-sequence currents, we can
find the sequence currents in the windings as follows:
i = + +
+
a ,1 a
eq SS
ein
n nz
z zj
1 2 213
1
=+ +
+
ia ,2 a
eq SS
ein
n nz
z zj
2 2 213
1
= + +
+
ia ,0 aeq,0 SS
in
n n nz
z z0 203
1 1( )
,,0
(7.96)
Here, ia1, ia2, and ia0 are the fault sequence currents for the
various fault types described in Chapter 6. These can then be
transformed to the phase or wind-ing currents by using Equation
6.1.
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 181
7.7.2 Standard Delta Fault Currents
This transformer has three windings. From Equation 7.39, we
have
I I Ia S1 L1e=
=
=2
13
2
13
n jn
n jn
j +
232
2
nj
e L1I (7.97)
where
I IL1 S1e and tan= =
j
n 2 31 (7.98)
Converting to the per-unit system, we find for the ratio of
current bases:
II
EV
nn
b,L1
b,a
a
L1
= =+3
32 (7.99)
Using this ratio, Equation 7.97 becomes (in per-unit terms)
i ia L1e= +2 3
322
nn
j
(7.100)
We also have
I I I I = =a L1 a L1e ,j (7.101)
The ratio of current bases is the same for the currents in
Equation 7.101 and is given by
II
EV n
b,L1
b,a'
a
L1
= =+
332
(7.102)
Using this, Equation 7.101 becomes (in per-unit terms),
i i i i = +=
+a L1 a L1e ,
33
332 2n n
j (7.103)
Using trigonometric identities and the formula for from Equation
7.98, we can show that ia + ia + ia = 0.
2010 by Taylor and Francis Group, LLC
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182 Transformer Design Principles
For the zero-sequence currents, from Equations 7.49 and 7.50, we
get
I I I Ia,0 L1,0 a ,0 L1,0 a ,0 L1,0, ,= = = 2n
I I (7.104)
Putting these on a per-unit basis, using the same current bases
as for the positive-sequence case, we get
i i i i ia,0 L1,0 a ,0 L1,0 a ,0, ,= +=
+=
2 33
33
32 2n n nn2 3+
iL1,0 (7.105)
These zero-sequence per-unit currents also sum to zero, as
expected.Using Equation 7.88 and noting that ieq corresponds to
iL1, we get the
following expressions for the winding sequence currents:
i ia,1 a1eq SS
a,2e ,= + +
=
i
nn
zz z
j2 332
2 1
iin
nz
z z
in
ja2
eq SS
a,0 a0
e2 3
3
2 3
22 1
2
+ +
=
i++ +
=+
3
33
0
2
1
zz z
in
zz
j
eq,0 SS,0
a ,1 a1 ei eeq SS
a ,2 a2eq SS
, e+
= + +
z
in
zz z
ji3
321
=+ +
i
i
a ,0 a0eq,0 SS,0
a ,1
in
zz z
332
0
==+ +
= +
in
zz z
in
za1
eq SSa ,2 a2,
33
332
1
2
1izz z
in
zz z
eq SS
a ,0 a0eq,0 SS,0
+
=+ +
i
332
0
(7.106)
The per-unit currents for the various fault types can be
inserted for ia1, ia2, and ia0, as shown in Chapter 6. Accordingly,
one can then get the winding phase currents in per-unit terms by
applying Equation 6.1 for the various fault types.
7.7.3 Two-Core Phase-Shifting Transformer Fault Currents
We will quote the results for this transformer based on the
formulas in Section 7.5. This transformer has five windings,
labeled 1, 1, a, a, and a. The per-unit winding currents, expressed
in terms of the output L1 current, are given by
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 183
i i i i1 2 2 1 12 3
32 3=
( ) += = j
n n
n n
nje s
e sL a
ee
, nn
n n
n n
j
j
s
e sL
a
e s
L
e
e
( ) +
=( ) +
22 1
2 1
3
3
3
i
i i , ii i
i i
=( ) +
= = =
a
e s
L
a
e
3
3
02 3
2 1
1 0 1 0 0
n n
in
, , ,, nn
n n
s
L
a
e s
L a
( ) +
=( ) +
=
2 1 0
0 2 1 0 0
3
3
3
i
i i i
,
, , ,
(7.107)
The winding sequence currents are given by
i i i1 1 2 2 1 1 1 22 3
32
, , ,, = ( ) += j
n n
n ne j
je s
e sL
333
0
2
22 1 2
1 0 1 0
1
n n
n ne
i
jL
e s
e s
a
( ) += =
=
i
i
i
,
, ,
,33
32 3
22 1 1 2
n n
n ne
n n
n nje s
e sL a
e s
e s( ) +=
i i, ,, (( ) +
= ( ) +
=
22 1 2
0 2 1 0
1
3
2 3
3
e
n n
ji
i i
i
L
a
e s
L
a
,
, ,
,33
3
3
321 1 2 2 1n n
en n
ej j
e s
L a
e s
L( ) +
=( ) +
i i i, ,, ,,
, ,
,
2
0 2 1 0
1 2
3
3
3
3
i i
i i
=( ) +
=( ) +
a
e s
L
a
e s
L
n n
n n 11 1 2 0 2 1 0
33, , , ,
,( )
= =+
i i ia ae s
L n n
(7.108)
Substituting the ieq currents from Equation 7.88 for iL1 into
the formulas in Equation 7.108, we get the sequence currents in all
the windings. Note that j in Equation 7.108 is a rotation by 90 in
the complex plane, so in the negative-sequence current formula, it
has to be replaced by j for a 90 rotation in the opposite
direction.
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184 Transformer Design Principles
7.8 Zig-Zag Transformer
A delta HV (H) winding and zig-zag low-voltage (X) winding
transformer is analyzed. The complex currents are calculated, as
well as the terminal positive- or negative- and zero-sequence
impedances. Although explicit expressions for these impedances are
given in terms of the two-winding impedances, they can also be
obtained directly from a program that calcu-lates the magnetic
field associated with complex currents. These impedances can then
be used to obtain fault currents for a short-circuit analysis.
The three-phase connections for the delta primary and zig-zag
second-ary windings are given schematically in Figure 7.12.
Parallel lines denote the windings on the same phase. The delta
winding phases are denoted a, b, and c. The zig-zag windings
consist of windings from the two phases connected in series. Thus,
the zig-zag X1 terminal is attached to a series combination of
windings from the b and c phases. The orientation of the zig-zag
windings in Figure 7.12 is such that the terminal voltage of the X
terminals to ground is in phase with the terminal voltage of the
delta ter-minals V to ground. This requires the two separate
windings of the zig-zag to have the same number of turns. A 180
phase shift between X and V can be achieved by winding the X
windings in the opposite sense of the delta windings.
Figure 7.13 shows the positive-sequence phasors associated with
the wind-ing and terminal voltages of the delta winding. Ea, Eb,
and Ec are a positive-sequence set of winding voltages, and V1, V2,
and V3 are a positive-sequence set of terminal-to-ground voltages.
The Es are no-load voltages due to the core excitation; the diagram
in Figure 7.13 does not take into account the voltage drops caused
by the currents and impedances. For the voltage drops,
a
b cc
b
ab c
a
V1
V2V3
X1
X2X3
Figure 7.12Schematic three-phase connection diagram of the
zig-zag transformer.
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 185
we use a three-winding T-equivalent circuit model. The single-
winding impedances are the same for all the three phases.
Concentrating on the a phase, the winding voltages are given by
V E I
V E Ia a a a
a a a a
Delta HVZig-Y LV
= += +
Z
Z
( )( ))( )V E I = +a a a a Zag-Y LVZ
(7.109)
where the currents are taken to be positive into the winding
terminals. The Es are proportional to the winding turns and
increase in the opposite sense to the current direction.
The connection diagram of the winding for all the three phases
is shown in Figure 7.14. The subscripts a, a, and a refer to the
delta, zig, and zag wind-ings, respectively, of the a phase. The b
and c subscripts have the same sig-nificance for the b and c
phases. The zig and zag windings are connected in such a way that
their phase voltages are subtractive. This gives the maxi-mum net
voltage across both windings.
7.8.1 Calculation of electrical Characteristics
Like the voltages, the currents I1, I2, and I3 and Ia, Ib, and
Ic form three-phase sets, as shown in Figure 7.14. From this
figure, we have
I I I
I I I
I I I
1
2
3
= = =
c b
a c
b a
(7.110)
Ea
Eb
Ec
V2
V1
V3
Figure 7.13Phasor diagram of the delta terminal and winding
voltages.
2010 by Taylor and Francis Group, LLC
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186 Transformer Design Principles
From Figure 7.14, we also have
V V E I
V V E I
V V E I
1 2
3 1
2 3
= + = + = +
c c a
b b a
a a a
Z
Z
Z
(7.111)
Note that Za = Zb = Zc; similar is the case for Za and Za.
Therefore, we will only use the subscripts a, a, and a for the
Zs
For the X1 terminal, from Figure 7.14, we see that
V E I E IX1 c c a b b a= + Z Z (7.112)
We let N1 = the number of turns in the HV delta winding and N2 =
the num-ber of turns in the zig winding = the number of turns in
the zag winding; thus, we have
EEEE
a
a
a
a
= =
=
NN
n12
1
(7.113)
where we define n as the ratio of turns in the delta winding to
turns in the zig or zag winding. Because of the phase relationship
among the constitu-ents of a positive-sequence set, we can
write
V1
V2V3
VX1
VX3
VX2
Ib
Ea+ Ia
Eb+
Ic
Ec+
IaIa
EaEa
IcIc
EcEcIb
Ib
EbEb
I3 I2
I1
Figure 7.14Detailed connection diagram of the three-phase
zig-zag transformer, with the electrical quan-tities labeled.
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 187
E E E E E = ( ) = =c b a a ae ej j j j n120 120 3
3 (7.114)
From Equation 7.111, we have
E V V I V Ia a a a ae e= ( ) = ( ) = 2 3 120 120 1 3Z Z jj j VV
I1 a aZ (7.115)
Substituting from Equations 7.114 and 7.115 into Equation 7.112,
we get
V V I I IX1 a a c a b a= + 3 3
1nj
nZ Z Z (7.116)
Due to the continuity of the current in the zig and zag
windings, we must have
I I
I I
I I
= = =
c b
a c
b a
(7.117)
so that Equation 7.116 can be rewritten as
V V I IX1 a a c a a= + +( ) 3 31n j n Z Z Z (7.118)
Since the currents form a balanced set, we have
I I =c ae j120 (7.119)
From amp-turn balance, we have
N N Nn
1 2 2 00
I I I
I I Ia a a
a a a
+ + =+ + =
(7.120)
From Equation 7.117, we have
I I Ia b ae = = j120 (7.121)
Thus, from Equations 7.119 through 7.121, we get
I I = c ajn3
(7.122)
Using this result, Equation 7.118 becomes
V V IX1 a a a a= + +( )
3 331
2
nj
nZ
nZ Z (7.123)
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188 Transformer Design Principles
From Equation 7.110, we have
I I I1 120 120 3= ( ) = e e a aj j j (7.124)
Substituting this into Equation 7.123, we get
V VI
X1 a a a= + +( )
331
12
n nZ
nZ Z (7.125)
Thus, under no-load conditions, we see from Equation 7.125 that
the terminal voltages are related by
V VX1 1=3n
(7.126)
This shows that the no-load terminal voltages H and X are in
phase, as desired. From Equations 7.122 and 7.124, the current into
the X1 terminal is
I I IX1 c 1= = n3
(7.127)
Thus, we see that the power into the X1 terminal, ignoring the
impedance drop, is
V I V IX1 X1 1 1* *= (7.128)
where * indicates complex conjugation. This equation shows that
the power flows out of the X1 terminal and that it equals the power
into the V1 terminal. From Equations 7.115 and 7.124, ignoring the
impedance drop, we also get
E I V Ia a* *= 1 1 (7.129)
so that the delta winding power equals the input power. The
a-phase wind-ing currents can be obtained from Equations 7.120 and
7.121.
I I
I I
= +
=
a a
a a
nj
nj
12 2 3
12 2 3
(7.130)
Multiplying Equation 7.125 by n/3, we get
n Zn
Z Z3 3 31
12
V VI
X1 a a a= + +( )
(7.131)
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 189
Thus, as seen from the high-voltage (HV) side, the effective
impedance, Zeff, is given by
Z Zn
Z Zeff a a a= + +( )
13 3
2
(7.132)
7.8.2 Per-unit Formulas
At this point, it is desirable to represent everything on
per-unit bases. We will use the input power per phase as the power
base. Using Equation 7.129, this also equals the delta winding
power. Thus, Pb = V1I1 = EaIa. The aster-isks are not needed
because these quantities are real. Thus, V1 and I1 are the voltage
and current bases for the H terminal. We label these as Vb,1 and
Ib,1. In terms of these and from Equations 7.126 and 7.127, the
base voltage and current at the X1 terminal are
Vn
V
In
I
b,X1 b,1
b,X1 b,1
=
=
3
3
(7.133)
Dividing Equation 7.131 by Vb,1 and using Equation 7.133, we
find that
v v iX1 eff= 1 1z (7.134)
where lowercase letters have been used for per-unit quantities,
and
zZZeff
eff
b,1
= (7.135)
where
ZVI
VPb,1
b,1
b,1
b,12
b
= = (7.136)
Since the H winding power equals the terminal power, we can
resort to a winding base to calculate zeff. From Equations 7.115
and 7.124, ignoring impedance reductions, we get
V V
I I
ZVI
VP
b,a b,1
b,a b,1
b,ab,a
b,a
b,a2
b
=
=
= = =
3
13
33Zb,1
(7.137)
2010 by Taylor and Francis Group, LLC
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190 Transformer Design Principles
For the a and a windings, using Equation 7.113 and maintaining
the same power base, we have
V V
Vn
ZVP
Z
b,a b,ab,a
b,a b,ab,a
b
b,
= =
= = =Z2
aa
n2
(7.138)
From Equations 7.135 and 7.132, we can write
zZZ
ZZ
ZZ
n Zeff
eff
b
b a
b
a
b a
b= =
+
,
,
, ,
1 1
213 3
,,
,
,
, ,
ab
b a
b a
a a
bZZZ
Z ZZ1
+( )
aa
(7.139)
Using Equations 7.137 and 7.138, we can put each winding on its
own base to get
z z z zeff a a a= + +( ) 13 (7.140)
In terms of two-winding per-unit impedances, we have the
standard formulas
zz z z
zz z z
aaa aa a a
aaa a a aa
= +
= +
2
2
zzz z z
= + a aa a a aa2
(7.141)
Substituting these into Equation 7.140, we get
zz z z
effaa aa a a= +
2 6 (7.142)
Thus, the effective terminal impedance can be derived from the
two- winding impedances.
7.8.3 Zero-Sequence impedance
We can use a similar analysis to get the zero-sequence
impedance, using a zero subscript to label the zero-sequence
quantities. Thus, instead of Equation 7.112, we have
V E I E IX1,0 c ,0 c ,0 a ,0 b ,0 b ,0 a ,0= + Z Z (7.143)
2010 by Taylor and Francis Group, LLC
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Phase-Shifting and Zig-Zag Transformers 191
However, Ec,0 and Eb,0 are now in phase and have the same
magnitude, and therefore, they cancel each other in Equation 7.143.
In addition, using Equation 7.117, which also applies to
zero-sequence currents, Equation 7.143 becomes
V IX c a a1 0 0 0 0, , , ,= +( ) Z Z (7.144)From Equation 7.117
and the equality of the zero-phase currents in the three phases, we
have
I I I = = a b a, , ,0 0 0 (7.145)
Hence, applying Equation 7.120 to zero-phase currents, we
get
Ia,0 = 0 (7.146)
Thus, no zero-sequence current flows in the delta HV winding.
Since IX1,0 = Ic,0, Equation 7.144 becomes
V IX1,0 X1,0 a ,0 a ,0= +( ) Z Z (7.147)Reverting to per-unit
quantities, we divide Equation 7.147 by Vb,X1 to get
v iX1,0 X1,0 eff,0= z (7.148)
where
zZ Z
Zeff,0a ,0 a ,0
b,X1
=+
(7.149)
and
ZV
PVV
VP
Zb,X1b,X12
b
b,X1
b,a
b,a2
b
= =
=
2
3 bb,a (7.150)
Substituting this into Equation 7.149 and using Equation 7.141,
we get
z z z zeff,0 a ,0 a ,0 a a ,0= +( ) = 1313
(7.151)
Thus, we can find the effective zero-sequence impedance from the
per-unit two-winding zero-sequence impedance between the zig and
zag windings. Normally, we calculate the positive-sequence
impedance and adjust it, via an empirical factor, to get the
zero-sequence impedance.
2010 by Taylor and Francis Group, LLC
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192 Transformer Design Principles
7.8.4 Fault Current analysis
The phase-sequence diagrams are shown in Figure 7.15. We will
calculate the sequence currents in the three windings in terms of
the fault-sequence currents, which depend on the fault type. The
phase currents can then be obtained by the transformation given in
Equation 6.1. We will consider equal positive- and
negative-sequence reactances for the transformer and the sys-tems
and will not distinguish them with subscripts. In addition, we will
con-sider an X-terminal fault, which is fed from the HV line.
The Thevenin impedances, in terms of those given in Figure 7.15,
are
z zz z zz z z
zz z
1 2 00= =
+( )+ +
=SX eff SHeff SH SX
SX ef, , ff SH
eff SH SX
, ,
, , ,
0 0
0 0 0
+( )+ +
z
z z z (7.152)
iX,2
iX,0
ia2
ia0
ZSX
ZSX,0
ZSH
eSH eSX
iX,1
ia1
ZSXZef f
Zef f
Zef f,0
ZSH
ZSH,0
(c)
(b)
(a)
Figure 7.15Sequence circuits for the zig-zag transformer: (a)
positive sequence, (b) negative sequence, and (c) zero
sequence.
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Phase-Shifting and Zig-Zag Transformers 193
zSH,0 must be set equal to zero in the formulas. The per-unit
sequence cur-rents feeding the fault from the transformer side are
given by
i iz
z zi i
zz zX,1 a1 eff SH
X,2 a2eff S
,=+
=+
1 1HH
X,0 a0eff,0 SH,0
=+
i i
zz z
0 (7.153)
We need to find the per-unit fault-sequence currents in the
windings in terms of the fault-sequence currents out of the X1
terminal. The number 1 preceded by a comma in Equation 7.153 refers
to a positive sequence, and so forth, whereas X1 refers to the X
terminal number 1. From Equations 7.124, 7.127, and 7.130, we find
that
I I I I I Ia X1 a a a a, ,= = +
= j nn j n
3 12
12 3
12
j1
2 3 (7.154)
From Equations 7.133, 7.137, and 7.138, the current bases are
related by
II
n II
II
b,X1
b,a
b,X1
b,a'
b,X1
b,a
,= = =3
13
(7.155)
Using these base-current relations, Equation 7.154 can be
written in per-unit terms as
i i i i i ia X1 a X1 a X1, ,= = +
= j j j j12
12 3
12
j1
2 3 (7.156)
The three per-unit currents sum to zero as expected.Replacing X1
by X,1, X,2, and X,0 for the sequence currents obtained from
Equation 7.153 for the different fault types, we can obtain,
from Equation 7.156, the fault-sequence currents in the various
windings. From these, we can then obtain the phase currents from
Equation 6.1.
2010 by Taylor and Francis Group, LLC
Chapter 7 Phase-Shifting and Zig-Zag Transformers7.1
Introduction7.2 Basic Principles7.3 Squashed Delta Phase- Shifting
Transformer7.3.1 Zero- Sequence Circuit Model
7.4 Standard Delta Phase- Shifting Transformer7.4.1 Zero-
Sequence Circuit Model
7.5 Two- Core Phase- Shifting Transformer7.5.1 Zero- Sequence
Circuit Model
7.6 Regulation Effects7.7 Fault Current Analysis7.7.1 Squashed
Delta Fault Currents7.7.2 Standard Delta Fault Currents7.7.3 Two-
Core Phase- Shifting Transformer Fault Currents
7.8 Zig- Zag Transformer7.8.1 Calculation of Electrical
Characteristics7.8.2 Per- Unit Formulas7.8.3 Zero- Sequence
Impedance7.8.4 Fault Current Analysis